Question

X=acos^3t，y=asin^3t 所 围成的平面图形的面积

请把详细过程写下，谢谢!用定积分

Answer 1

x=acos^3t,y=asin^3t是星形线，它的面积为
∫ydx=4*∫asin^3t(acos^3t)'dt,t:π/2→0
=-3*a^2∫sin^4t*cos^2tdt
=-3a^2∫(sin^4t-sin^6t)dt
=3/8*πa^2

Answer 2

y=[a^(2/3) - x^(2/3)]^(3/2) （设a>0)
S=4 定积分(0,a) ydx, 令x=a^(2/3) cos^2 t
S=4 定积分 -2/5 a^(2/3) sin^5 t
=8/5 a^(2/3)(1-a^(2/3) )^5

Answer 3

S = |2∫ydx| (t=0到t=π) = |2∫asin^3t * a*3cos^2t*(-sint)dt| = 6aa∫sin^4tcos^2tdt
= 6aa/5 ∫sin^6tdt
=3aa/20π
只能输100字，没办法了